16 \( \begin{array}{l}\frac{3}{5} \text { of the pupils in Primary } 6 A \text { and } \frac{1}{2} \text { of the pupils in Primary } 6 B \text { are girls. } \\ \text { Both classes have the same number of girls. Primary } 6 B \text { has } 6 \text { move boys } \\ \text { than Primary } 6 A \text {. } \\ \text { a) How many pupils are there in Primary } 6 B \text { ? } \\ \text { b) } \frac{1}{3} \text { of the girls and } \frac{2}{5} \text { of the boys in the two classes wear spectacles. } \\ \text { How many pupils in the two classes wear spectacles altogether? }\end{array} \)

Let's solve the problem step by step. ### **Given:** 1. **Primary 6 A:** - \( \frac{3}{5} \) of the pupils are girls. 2. **Primary 6 B:** - \( \frac{1}{2} \) of the pupils are girls. 3. Both classes have the **same number of girls**. 4. **Primary 6 B** has **6 more boys** than **Primary 6 A**. ### **Let’s Define Variables:** - Let \( G \) be the number of girls in each class. - Let \( A \) be the total number of pupils in Primary 6 A. - Let \( B \) be the total number of pupils in Primary 6 B. ### **Using the Information:** 1. **For Primary 6 A:** \[ \frac{3}{5}A = G \quad \Rightarrow \quad \text{Number of boys in 6 A} = A - G = A - \frac{3}{5}A = \frac{2}{5}A \] 2. **For Primary 6 B:** \[ \frac{1}{2}B = G \quad \Rightarrow \quad \text{Number of boys in 6 B} = B - G = B - \frac{1}{2}B = \frac{1}{2}B \] 3. **Equal Number of Girls:** \[ \frac{3}{5}A = \frac{1}{2}B \quad \Rightarrow \quad B = \frac{6}{5}A \] 4. **Difference in Number of Boys:** \[ \frac{1}{2}B - \frac{2}{5}A = 6 \] Substitute \( B = \frac{6}{5}A \): \[ \frac{1}{2} \times \frac{6}{5}A - \frac{2}{5}A = 6 \quad \Rightarrow \quad \frac{3}{5}A - \frac{2}{5}A = 6 \quad \Rightarrow \quad \frac{1}{5}A = 6 \quad \Rightarrow \quad A = 30 \] 5. **Total Pupils in Primary 6 B:** \[ B = \frac{6}{5} \times 30 = 36 \] ### **Answer to Part (a):** **Primary 6 B** has **36 pupils**. ### **For Part (b):** 1. **Total Girls in Both Classes:** \[ G_{\text{total}} = 18 + 18 = 36 \quad \left(\text{since } \frac{3}{5} \times 30 = 18 \text{ and } \frac{1}{2} \times 36 = 18\right) \] 2. **Total Boys in Both Classes:** \[ B_{\text{total}} = 12 + 18 = 30 \quad \left(\text{since } \frac{2}{5} \times 30 = 12 \text{ and } \frac{1}{2} \times 36 = 18\right) \] 3. **Pupils Wearing Spectacles:** - \( \frac{1}{3} \) of the girls: \[ \frac{1}{3} \times 36 = 12 \] - \( \frac{2}{5} \) of the boys: \[ \frac{2}{5} \times 30 = 12 \] - **Total pupils wearing spectacles:** \[ 12 + 12 = 24 \] ### **Answer to Part (b):** A total of **24 pupils** in the two classes wear spectacles. ## **Final Answers:** a) **36 pupils** in Primary 6 B. b) **24 pupils** wear spectacles altogether.